Which point on x-axis is equidistant from (5, 9) and (-4, 6)? - Vidyadip

Question

Which point on x-axis is equidistant from (5, 9) and (-4, 6)?

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Answer

Let A(5, 9) and B(-4, 6) be the given points. Let C(x, 0) be the point on x-axis Now,$\text{AC}=\sqrt{(\text{x}-5)^2+(0-9)^2}$
$\Rightarrow\ \text{AC}=\sqrt{\text{x}^2+25-10\text{x}+(-9)^2}$
$\Rightarrow\ \text{AC}=\sqrt{\text{x}^2-10\text{x}+25+81}$
$\Rightarrow\ \text{AC}=\sqrt{\text{x}^2-10\text{x}+106}$
$\text{BC}=\sqrt{(\text{x}+4)^2+(0-6)^2}$
$\Rightarrow\ \text{BC}=\sqrt{\text{x}^2+16+8\text{x}+(-6)^2}$
$\Rightarrow\ \text{BC}=\sqrt{\text{x}^2+8\text{x}+16+36}$
$\Rightarrow\ \text{BC}=\sqrt{\text{x}^2+8\text{x}+52}$
$Since, AC = BC Or, AC^2 = BC^2 x^2 - 10x + 106 = x^2 + 8x + 52$
$\Rightarrow -10x + 106 = 8x + 52 \Rightarrow -10x - 8x = 52 - 106$
⇒ -18x = -54 $\Rightarrow\ \text{x}=\frac{54}{18}$
⇒ x = 3 Hence the points on x-axis is (3, 0).

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